2 Delaunay Triangulations

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1 and the closed half-space H(b, a) containingb and with boundary the bisector hyperplane is the locus of all points such that (b 1 a 1 )x 1 + +(b m a m )x m (b b 2 m)/2 (a a 2 m)/2. The closed half-space H(a, b) isthesetofallpointswhosedistancetoa is less than or equal /651: to the distancealgorithms to b, and vice versa for H(b, a). CMU, Spring 2015 Lecture Figure #17: 8.2 shows Voronoi the Diagrams Voronoi diagram and Delaunay of a set of Triangulations twelve points. March 23, 2015 Lecturer: Danny Sleator 1 Voronoi Diagrams Figure 8.2: A Voronoi diagram Given a set of sites (points) in the plane, s 1, s 2,... s n the Voronoi diagram partitions the plane into regions where the region associated with s i is the set of points in the plane that are closer to site s i thaninany theother. general Letcase R i denote wherethe E region has dimension associatedm, withthe s i. definition The diagram of above the Voronoi shows adiagram Voronoi diagram Vor(P )ofp with twelve is thesites. same as Definition 8.1, except that H(p i,p j )istheclosedhalf-space The containing Voronoip regions i and having are convex the bisector polygons. hyperplane To see thisof note a and thatb Ras i is boundary. the intersection Also, observe of n 1 halfplanes, the convex one for hulleach of Pother is a site convex j. It s polytope. the half space of the plane (bounded by the perpendicular that bisector We will between now sstate i and as j lemma ) of the listing points closer the main to s i properties than to s j. of SoVoronoi the Voronoi diagrams. regions are It turns (possibly out infinite) that certain convex degenerate polygons. situations Because the canvoronoi be avoided diagram if we is assume a planarthat graph, if Ptheis number a set of of points vertices in and edges in it is linear in n (by virtue of Euler s formula). an affine space of dimension m, then no m +2pointsfromP belong to the same (m 1)- We sphere. will talk We about will say Voronoi that the regions points (mentioned of P areabove), in general Voronoi position. edges (boundaries Thus whenbetween m =2,no3.5 Voronoi regions) points inand P are Voronoi cocyclic, vertices and(where whenedges m =3,no5pointsinP meet other edges). are on the same sphere. When we say that two Voronoi regions have a boundary in common or share a boundary we mean a boundary of non-zero length. So, for example, if there are four sites that are at the corners of a square, then the vertex of the Voronoi diagram has degree four. We don t consider diagonally opposite regions to have a boundary in common. Unless otherwise specified, in these notes we re going to assume that the Voronoi diagram has only vertices of degree three. (This will happen if no four of the sites are co-circular.) 1

2 2 Delaunay Triangulations A triangulation of a set of sites is a way of partitioning the convex hull of the set of sites into triangles, where the vertices of the triangles consist of sites. The Delaunay triangulation of a set of sites is a specific triangulation that is the dual of the 152 Voronoi diagram. Two sites s i and s j are CHAPTER connected by 8. andirichlet VORONOI edge in the Delaunay triangulation DIAGRAMS if and only if R i and R j share a boundary in the Voronoi diagram. The following diagram superimposes the Delaunay triangulation on the Voronoi diagram above. The Delaunay triangulation has a number of nice properties. Here are some of them. * The closest Figurepair 8.4: isdelaunay an edge of triangulation the Delaunayassociated triangulation. with(the a Voronoi following diagram figure, in conjunction with the lemma Proof below proves (closest this.) pair property) The most direct method to obtain fast algorithms is to use the lifting method discussed in Section 8.4, whereby the original set of points is lifted onto a paraboloid, and to use fast algorithms for finding a convex hull. empty s i empty A very interesting (undirected) graph can be obtained sj from the Voronoi diagram as follows: The vertices of this graph are the points p i (each corresponding to a unique region of Vor(P )), and there is an edge between p i and p j iff the regions V i and V j share an edge. Theresulting graph is called adelaunay triangulation of the convexhullof P, after Delaunay, who invented this concept in Such triangulations have remarkable properties. * Figure The minimum 8.4 shows spanning the Delaunay tree of triangulation the sites (usingassociated the Euclidean withdistance the earlier as the Voronoi lengthdiagram of an of a set edge) of is twelve a subset points. of the edges of the Delaunay triangulation. NUS CS4235 Lecture 9: Voronoi diagrams and Delaunay triangulations p.35/42 * One The has smallest to be angle careful in to a triangulation make sure that is the all minimum the angle vertices among all have of the been triangles computed in it. The Delaunay triangulation, among all triangulations, has the maximum smallest angle. before computing a Delaunay triangulation, since otherwise, some edges could be missed. In Figure * You 8.5can illustrating use the Delaunay such a situation, triangulation if the to compute lowest Voronoi the Voronoi vertex diagram. had not been computed (not shown on the diagram!), the lowest edge of the Delaunay triangulation would be missing. 2 m. The concept of a triangulation can be generalized to dimension 3, or even to any dimension

3 Lemma: The edges of the convex hull of the sites are edges in the Delaunay triangulation. Proof: If two sites are neighbors on the convex hull, then if we follow the perpendicular bisector between the points sufficiently far away, eventually we must reach a point which is closest to the two neighbors than any other site. This shows that the two Voronoi regions have this boundary incommon. QED. Proof (Empty circle property) Lemma: (s i, s j ) is in the Delaunay triangulation iff there exists a circle through s i and s j containing no other site inside of it. s i s j DT (S) Proof: : (s i, s j ) being in the Delaunay triangulation means that R i and R j have a boundary in common. Take a circle centered on a point on that boundary. This circle cannot contain another site inside of it (this would violate the fact that this point is closer to s i and s j than any other site.) : There exists a circle through s i and s j containing no other site. Let p be the center of this circle. Point p must be on the boundary between R i and R j. If p is not a vertex of the Voronoi diagram then it must be along an edge of the diagram, and we re done. If it is a vertex, then one NUS CS4235 Lecture 9: Voronoi diagrams and Delaunay triangulations p.33/42 of the three lines eminating from that vertex is the boundary between R i and R j. QED. Lemma: Assuming no four sites are co-circular, then the Deleunay triangulation is unique. Proof: Every vertex of the Voronoi diagram has degree 3. These correspond to the triangles of the Delaunay triangulation. QED. Note: Furthermore if there exists four or more co-circular sites, the Delaunay triangulation is not unique. It is unique if you allow the Delaunay triangulation to have non-triangulated polygons (all of whose vertices are co-circular.) Claim: Given a Delaunay triangulation for a set of sites we can compute the Voronoi diagram in O(n) time. And conversely. 3

4 3 Computing the Delaunay Triangulation 3D-Convex Hull Method: Take each site s i = (x i, y i ) and replace it by a point in three dimensions (x i, y i, x 2 i + y2 i ). This is a point on a parabaloid of revolution. Now take the convex hull of this set of points in 3D. The triangles of this convex hull (visible from the x, y plane), when projected back down to the plane, form the the Delaunay triangulation. No Proof. Note that the 3D convex hull can be computed in O(n log n) time. The following figure illustrates this process. Voronoi diagram: A different Formulation 1. Project each point p i on the surface of a unit paraboloid Today I will present a simple algorithm for computing the Delaunay triangulation in O(n 2 ) time. Again we assume that no three sites are co-circular. (Although this assumption can easily be removed if we allow the algorithm to output co-circular polygons, instead of only triangles.) 2. Compute the lower convex hull of the projected points. Simple Delaunay Algorithm: Result: Given S = {p i i=1, 2, n} in the plane (no 4 points co-circular) and given 3 points (1) Findp, anq, edge r S, thatthe mustriangle be in the Delaunay pqr is a triangulation. triangle of (The Delauney closest pair triangulation has his if property, as well as any edge of the convex hull of the sites.) p q r is a face of the lower of the projected points S We re going to maintain a queue of directed Delaunay edges that need to have the triangle Conclusion: to their The right projection be explored. of Put this the convex starting edge hull just gives found the (both Delauney directions oftriangulation it) into this queue to start with, and also add them to the Delaunay triangulation we re building. of the point set. (2) Process the queue in the following way until the queue is empty. Let (i, j) be an edge in the queue. Consider all the other sites k such that s k is to the right of ray s i s j. If there is no such k, then (i, j) is a convex hull edge. Delete it from the queue. Otherwise, find the s k to the right of s i s j so that the circle (s i, s j, s k ) does not contain any other sites. We do this by simply trying them all, and keeping the best one found so far. If a new one is inside the circle of the best previously found one, this one becomes our new best. (This requires the incircle test described in an earlier lecture.) Now, having found our best site s k, we add edges (i, k), (k, j) to the queue, and to the Delaunay triangulation we re building. We also remove (i, j), (j, k), and (k, i) from the queue. 4

5 A couple of comments: * Each edge generated is a Delaunay edge. The first one (the closest pair) clearly is just draw the circle with this edge as its diameter. It must contain no other sites. Every additional edge that we add comes with an associated empty circle that certifices it as a Deleunay OneNote Online edge. * The circle through the points (s i, s j, s k ) found in the search above must be an empty! circle. We know "#$%&'(!)&#*+!,-(!,./0! this 01.2!3)! because we have inductively certified that (s i, s j ) is a Delaunay edge. Specifically the following situation cannot arise: There cannot be a site (such as s l shown above) on the left side of the ray s i s j, and inside the circle. The existance of such a site proves that there is NO empty circle through sites s i and s j. And we know that is impossible by virtue of our proof that (s i, s j ) is a Delaunay edge. * When the algorithm stops, the entire triangulation has been generated. This follows from the fact that the triangulation is connected. * The running time is O(n 2 ) because for each Delaunay edge generated we may have to do a scan of all the other sites. 4 An Additional Links A lecture by Ken Clarkson: This applet lets you insert, delete, and move sites around while viewing the Delaunay triangulation 1 of 1 3/28/15 12:18 AM and/or the Voronoi diagram: Here s my Ocaml implementation of the O(n 2 ) algorithm described above: 5